Formula
Latent heat
Q = m L during phase transition at constant T. L_fusion (water): 3.34 × 10⁵ J/kg. L_vaporisation (water): 2.26 × 10⁶ J/kg. Latent heat is absorbed/released without temperature change.
-- NCERT Class 11 Physics, Ch. 10, p. 213Change of State Latent Heat
8 MCQs4 revision cards9-step worked example
Source: NCERT Properties of Solids and LiquidsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Change of State Latent Heat, explained for NEET
A common confusion in NEET problems on change of state: treating latent heat as "extra temperature" instead of recognising that temperature stays constant during a phase transition. The heat you supply at the melting or boiling point does not raise temperature — it breaks intermolecular bonds.
What latent heat means. When a substance reaches its melting point (or boiling point), additional heat goes entirely into changing the phase. The formula is Q = mL, where L is the specific latent heat — either L_fusion (solid → liquid) or L_vaporisation (liquid → gas). For water: L_fusion ≈ 3.34 × 10⁵ J/kg; L_vaporisation ≈ 2.26 × 10⁶ J/kg (NCERT Class 11 Physics, Chapter 10, page 208).
The flat region on the heating curve. Plot temperature vs heat supplied for ice → water → steam. You get two flat plateaus — one at 0 °C (fusion) and one at 100 °C (vaporisation). During each plateau, Q = mL applies; between plateaus, Q = mcΔT applies.
Where NEET problems test you. A typical question gives you a mass of ice at some sub-zero temperature and asks for the total heat to convert it fully to steam at 100 °C. You must chain three or more stages: (1) heat ice from initial temperature to 0 °C using Q = mcΔT with c_ice, (2) melt ice at 0 °C using Q = mL_fusion, (3) heat water from 0 °C to 100 °C using Q = mcΔT with c_water, (4) vaporise water using Q = mL_vaporisation. Missing any stage — or applying the wrong specific heat for the wrong phase — costs you the mark.
Watch out: L_vaporisation is roughly 6.8 times L_fusion for water. If your answer for total heat is dominated by the fusion step rather than the vaporisation step, re-check — you may have swapped the two values.
Can you answer these Change of State Latent Heat MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
During the melting of ice at 0 °C and 1 atm pressure, the temperature of the ice-water mixture:
Show answer and why every option is right or wrong
Answer: C. During a phase change (melting), all supplied heat is used to overcome intermolecular forces. Temperature remains constant at the melting point until the entire substance has changed phase (NCERT Class 11 Physics, Chapter 10, page 208).
Why A is wrong: A is wrong because temperature does not increase during a phase change — the heat supplied goes into breaking bonds, not raising kinetic energy of molecules.
Why B is wrong: B is wrong because the system absorbs energy during melting, but this energy changes the phase, not the temperature. There is no temperature decrease.
Why D is wrong: D is wrong because there is no fluctuation at the melting point under constant pressure. The temperature is strictly constant until the phase change is complete.
MCQ 2Easy RecallPractice
The SI unit of specific latent heat is:
Show answer and why every option is right or wrong
Answer: C. Specific latent heat L is defined as heat per unit mass for a phase change: Q = mL, so L = Q/m. Units: J/kg (NCERT Class 11 Physics, Chapter 10, page 208).
Why A is wrong: A is wrong because J/(kg·K) is the unit of specific heat capacity c, not latent heat. Latent heat has no temperature term since temperature doesn't change during a phase transition.
Why B is wrong: B is wrong because J/K is the unit of heat capacity (total, not per unit mass). Latent heat is defined per unit mass, not per degree.
Why D is wrong: D is wrong because J·kg has no physical meaning in this context. L = Q/m gives J divided by kg, not J multiplied by kg.
MCQ 3Easy RecallPractice
For water at standard pressure, the specific latent heat of vaporisation is approximately:
Show answer and why every option is right or wrong
Answer: B. L_vaporisation for water ≈ 2.26 × 10⁶ J/kg. This is significantly larger than L_fusion because vaporisation requires breaking nearly all intermolecular bonds (NCERT Class 11 Physics, Chapter 10, page 208).
Why A is wrong: A is wrong because 3.34 × 10⁵ J/kg is the specific latent heat of fusion of ice, not vaporisation. Confusing the two values is a common error — L_vaporisation is roughly 6.8 times larger.
Why C is wrong: C is wrong because 4.18 × 10³ J/(kg·K) is the specific heat capacity of liquid water, not a latent heat value. Note the different units — specific heat has a per-kelvin term.
Why D is wrong: D is wrong because 2.09 × 10³ J/(kg·K) is the specific heat capacity of ice, not a latent heat. This value governs temperature change in ice, not phase change.
MCQ 4Direct ApplicationPractice
How much heat is required to melt 0.50 kg of ice at 0 °C completely into water at 0 °C? (L_fusion = 3.34 × 10⁵ J/kg)
Show answer and why every option is right or wrong
Answer: D. Q = mL_fusion = 0.50 × 3.34 × 10⁵ = 1.67 × 10⁵ J. Since the ice is already at 0 °C, only the latent heat portion applies — no mcΔT stage is needed.
Why A is wrong: A is wrong because 6.68 × 10⁵ J comes from dividing L by the mass instead of multiplying: 3.34 × 10⁵/0.50 = 6.68 × 10⁵. Check: 0.50 × 3.34 × 10⁵ = 1.67 × 10⁵.
Why B is wrong: B is wrong because 1.13 × 10⁶ J corresponds to using L_vaporisation instead of L_fusion. The question asks for melting (fusion), not boiling (vaporisation).
Why C is wrong: C is wrong because 3.34 × 10⁵ J corresponds to melting 1.0 kg of ice, not 0.50 kg. This error comes from forgetting to multiply L by the actual mass.
MCQ 5Direct ApplicationPractice
200 g of ice at 0 °C is mixed with 200 g of water at 80 °C in a thermally insulated container. What is the final temperature of the mixture? (L_fusion = 3.34 × 10⁵ J/kg, c_water = 4.18 × 10³ J/(kg·K))
Show answer and why every option is right or wrong
Answer: B. B is correct. Heat available from the water cooling to 0 °C is 0.200 × 4180 × 80 = 66,880 J; melting all the ice needs 0.200 × 3.34 × 10⁵ = 66,800 J. Since 66,880 > 66,800, ALL the ice melts, and the 80 J left over warms the resulting 400 g of water by only 80/(0.400 × 4180) ≈ 0.05 °C. The mixture settles at 0 °C with no ice left (NCERT Class 11 Physics Chapter 10, page 203).
Why A is wrong: A is wrong because this assumes simple averaging of temperatures without accounting for latent heat. The 66,880 J from the hot water is almost entirely consumed by melting ice (66,800 J), leaving almost nothing to raise the temperature above 0 °C.
Why C is wrong: C is wrong because reaching 20 °C would require an additional Q = 0.400 × 4180 × 20 = 33,440 J beyond what is needed for melting. The hot water supplies only ~80 J surplus after fusion, nowhere near enough.
Why D is wrong: D is wrong because the available 66,880 J EXCEEDS the 66,800 J of fusion, so no ice survives. The margin is only 80 J, which is why the final temperature is barely above zero — but 'some ice remaining' would require the inequality to run the other way.
MCQ 6Direct ApplicationPractice
On a heating curve (temperature vs heat supplied) for a pure substance, the slope of the curve during a phase change is:
Show answer and why every option is right or wrong
Answer: A. During a phase change, temperature remains constant while heat is absorbed (Q = mL). On a T vs Q graph, this produces a horizontal line — slope = ΔT/ΔQ = 0.
Why B is wrong: B is wrong because a positive slope means temperature is rising with heat input. That occurs between phase changes (Q = mcΔT regions), not during them.
Why C is wrong: C is wrong because a negative slope would mean temperature decreases as heat is added, which violates energy conservation for a substance at its phase-change temperature under constant pressure.
Why D is wrong: D is wrong because an infinite slope would mean temperature jumps instantly with no heat input. This doesn't happen — a finite quantity Q = mL must be supplied for the phase change to complete.
MCQ 7CalculationPractice
Calculate the total heat required to convert 100 g of ice at −10 °C to steam at 100 °C. (c_ice = 2.09 × 10³ J/(kg·K), L_fusion = 3.34 × 10⁵ J/kg, c_water = 4.18 × 10³ J/(kg·K), L_vaporisation = 2.26 × 10⁶ J/kg)
Show answer and why every option is right or wrong
Answer: A. Four stages, m = 0.100 kg. (1) Ice −10 °C → 0 °C: Q₁ = 0.100 × 2090 × 10 = 2,090 J. (2) Melt ice: Q₂ = 0.100 × 3.34 × 10⁵ = 33,400 J. (3) Water 0 °C → 100 °C: Q₃ = 0.100 × 4180 × 100 = 41,800 J. (4) Vaporise: Q₄ = 0.100 × 2.26 × 10⁶ = 226,000 J. Total = 2,090 + 33,400 + 41,800 + 226,000 = 303,290 J = 3.03 × 10⁵ J.
Why B is wrong: B is wrong because 2.26 × 10⁵ J equals only the vaporisation step (Q₄). This error comes from calculating only the final phase change while ignoring the three preceding stages.
Why C is wrong: C is wrong because 3.01 × 10⁵ J likely omits the initial heating of ice from −10 °C to 0 °C (stage 1). That stage contributes 2,090 J — small but necessary for the correct total.
Why D is wrong: D is wrong because 2.61 × 10⁵ J likely omits the water-heating stage (Q₃ = 41,800 J). All four stages must be summed: heating ice + melting + heating water + vaporising.
MCQ 8CalculationPractice
A 50 g copper calorimeter (c_copper = 390 J/(kg·K)) contains 100 g of water at 30 °C. A 20 g piece of ice at 0 °C is dropped in. Find the final equilibrium temperature. (L_fusion = 3.34 × 10⁵ J/kg, c_water = 4.18 × 10³ J/(kg·K))
Show answer and why every option is right or wrong
Answer: B. Heat needed to melt ice: Q_melt = 0.020 × 3.34 × 10⁵ = 6,680 J. Let final temperature = T. Heat lost by water + calorimeter cooling from 30 °C to T: Q_lost = (0.100 × 4180 + 0.050 × 390)(30 − T) = (418 + 19.5)(30 − T) = 437.5(30 − T). Heat gained by melted ice warming from 0 °C to T: Q_gain_water = 0.020 × 4180 × T = 83.6T. Energy balance: 437.5(30 − T) = 6,680 + 83.6T → 13,125 − 437.5T = 6,680 + 83.6T → 6,445 = 521.1T → T ≈ 12.4 °C. First verify all ice melts: Q_available at T = 0 °C check → 437.5 × 30 = 13,125 J > 6,680 J, so yes, all ice melts, and the final temperature is about 12.4 °C.
Why A is wrong: A is wrong because 30 °C ignores the ice entirely. The ice absorbs a large amount of heat for melting (6,680 J) plus further heat to warm up, significantly cooling the system.
Why C is wrong: C is wrong because checking: Q_available from water + calorimeter cooling to 0 °C = 437.5 × 30 = 13,125 J, while only 6,680 J is needed to melt the ice. Since 13,125 > 6,680, all ice melts and the final temperature is above 0 °C.
Why D is wrong: D is wrong because 22 °C likely results from forgetting the latent heat of fusion — treating the ice as if it were already liquid water at 0 °C. The 6,680 J consumed by melting must be included.
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Change of State Latent Heat: quick recall before you leave
How do you solve a Change of State Latent Heat question? A worked example
- 1
Given
• Mass: m = 50 g = 0.050 kg• Initial temperature: T_i = −20 °C• Final temperature: T_f = 40 °C• c_ice = 2.09 × 10³ J/(kg·K)• L_fusion = 3.34 × 10⁵ J/kg• c_water = 4.18 × 10³ J/(kg·K)
- 2
Required
Total heat Q_total to go from ice at −20 °C to water at 40 °C.
- 3
Concept
Three stages: (1) heat ice from −20 °C to 0 °C, (2) melt ice at 0 °C, (3) heat water from 0 °C to 40 °C. Each stage uses a different formula or different value of c.
- 4
Formulas
• Stage 1: Q₁ = m × c_ice × ΔT₁• Stage 2: Q₂ = m × L_fusion• Stage 3: Q₃ = m × c_water × ΔT₃
- 5
Substitution
• Q₁ = 0.050 × 2090 × 20 = ?• Q₂ = 0.050 × 3.34 × 10⁵ = ?• Q₃ = 0.050 × 4180 × 40 = ?
- 6
Calculation
• Q₁ = 0.050 × 2090 × 20 = 2,090 J• Q₂ = 0.050 × 334,000 = 16,700 J• Q₃ = 0.050 × 4180 × 40 = 8,360 J• Q_total = 2,090 + 16,700 + 8,360 = 27,150 J
Note on exact values: the temperature intervals (20 K and 40 K) are exact by problem definition and do not limit significant figures. - 7
Final answer
Q_total ≈ 2.72 × 10⁴ J
The dominant contribution is the melting step (16,700 J, about 61% of the total), followed by heating water (8,360 J, about 31%), then heating ice (2,090 J, about 8%). If a problem involves vaporisation as well, expect the vaporisation term to dominate overwhelmingly. - 8
Common trap
Forgetting to use c_ice for the sub-zero heating stage and instead using c_water throughout. Since c_ice ≈ 0.50 × c_water, this error overestimates Q₁ by a factor of 2. Always check: which phase is the substance in during each stage?
- 9
Similar NEET-style question
A 200 g block of ice at −15 °C is placed in an insulated container with 300 g of water at 50 °C. Determine the final equilibrium temperature and state of the mixture. (Same constants as above.)
Approach: calculate heat available from water cooling to 0 °C, compare with heat needed to warm ice to 0 °C + melt it. If surplus heat remains, all ice melts and final T > 0 °C. If deficit, some ice remains at 0 °C.
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What to remember before solving Change of State Latent Heat questions
Which Change of State Latent Heat formulas do you need for NEET?
1 formula — click to collapse
Latent heat
Heat absorbed/released during phase change at constant T. L_fusion or L_vaporisation.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Q | heat | J |
| m | mass | kg |
| L | latent heat | J/kg |
Valid when
- Phase transition (constant T during)
- All mass m undergoes the transition
More in Properties of Bulk Matter: 5 exam traps and mistakes · 11 formulas · 5 question patterns from its other lessons.
Change of State Latent Heat questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 17 past-paper questions from Properties of Bulk Matter →
Sources
NCERT refs: Class 11 Physics Chapter 10, p.208
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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